Showing a matrix is diagonalizable and calculating powers

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  • #1
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Homework Statement


Consider the matrix A =
| 7 16 8|
|-1 0 -1|
|-2 -10 -3|

Show that A is diagonalizable. Find an invertible matrix P and diagonal matrix D and use the obtained result in order to calculate A^2 and A^3

Homework Equations


Determinant equation I suppose.
D is the diagonal matrix, like identity matrix but constructed from the eigenvalues. The order of the eigenvalues must match the columns in P precisely.

The Attempt at a Solution


Alright well I've done a huge chunk of this problem already. I found the eigenvalues to be 2, 3, and -1. Then I find my eigenvectors, and I come up with 2 per eigenvalue.

For λ = 2 I get [-2 1 0] and [-1 0 1]
For λ = 3 I get [-3 1 0] and [-1 0 1]
For λ = 1 I get [1 1 0] and [0 1 1]

So now I have to construct P from these. This is where I'm confused. Which eigenvectors from which eigenvalues do I use? I've tried several combinations to make AP = PD and I just can't do it. Everything comes out wrong. I suppose there's just something I'm not understanding. Please help.
 

Answers and Replies

  • #2
35,285
7,128


Homework Statement


Consider the matrix A =
| 7 16 8|
|-1 0 -1|
|-2 -10 -3|

Show that A is diagonalizable. Find an invertible matrix P and diagonal matrix D and use the obtained result in order to calculate A^2 and A^3

Homework Equations


Determinant equation I suppose.
D is the diagonal matrix, like identity matrix but constructed from the eigenvalues. The order of the eigenvalues must match the columns in P precisely.

The Attempt at a Solution


Alright well I've done a huge chunk of this problem already. I found the eigenvalues to be 2, 3, and -1. Then I find my eigenvectors, and I come up with 2 per eigenvalue.

For λ = 2 I get [-2 1 0] and [-1 0 1]
For λ = 3 I get [-3 1 0] and [-1 0 1]
For λ = 1 I get [1 1 0] and [0 1 1]

So now I have to construct P from these. This is where I'm confused. Which eigenvectors from which eigenvalues do I use? I've tried several combinations to make AP = PD and I just can't do it. Everything comes out wrong. I suppose there's just something I'm not understanding. Please help.
Since you have a 3 x 3 matrix and found three eigenvalues, you should not have two eigenvectors per eigenvalue, so you have definitely done something wrong.

Once you have that squared away, it's a good idea to check your eigenvalues and eigenvectors. If λ is an eigenvalue, and x is an eigenvector associated with that eigenvalue, it should be true that Ax = λx. If not, you did something wrong.

If you end up with three eigenvectors, form a matrix P with those eigenvectors as columns. It doesn't matter what order you put them in, but the order will determine which eigenvalue goes where in your diagonal matrix D.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966


Homework Statement


Consider the matrix A =
| 7 16 8|
|-1 0 -1|
|-2 -10 -3|

Show that A is diagonalizable. Find an invertible matrix P and diagonal matrix D and use the obtained result in order to calculate A^2 and A^3

Homework Equations


Determinant equation I suppose.
D is the diagonal matrix, like identity matrix but constructed from the eigenvalues. The order of the eigenvalues must match the columns in P precisely.

The Attempt at a Solution


Alright well I've done a huge chunk of this problem already. I found the eigenvalues to be 2, 3, and -1. Then I find my eigenvectors, and I come up with 2 per eigenvalue.

For λ = 2 I get [-2 1 0] and [-1 0 1]
For λ = 3 I get [-3 1 0] and [-1 0 1]
For λ = 1 I get [1 1 0] and [0 1 1]

So now I have to construct P from these. This is where I'm confused. Which eigenvectors from which eigenvalues do I use? I've tried several combinations to make AP = PD and I just can't do it. Everything comes out wrong. I suppose there's just something I'm not understanding. Please help.
[tex]\begin{bmatrix} 7 & 16 & 8 \\ -1 & 0 & -1 \\-2 & -10 & -3\end{bmatrix}\begin{bmatrix}-2 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix}2 \\ 2 \\ -6\end{bmatrix}[/tex]
so this is NOT an eigenvector.

[tex]\begin{bmatrix} 7 & 16 & 8 \\ -1 & 0 & -1 \\-2 & -10 & -3\end{bmatrix}\begin{bmatrix}-1 \\ 0 \\ 1\end{bmatrix}= \begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}[/tex]
so this is and eigenvector but with eigenvalue -1, not 2.

[tex]\begin{bmatrix} 7 & 16 & 8 \\ -1 & 0 & -1 \\-2 & -10 & -3\end{bmatrix}\begin{bmatrix}-3 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix}-5 \\ 3 \\ -4\end{bmatrix}[/tex]
so this is not an eigenvector.

[-1 0 1] is the same vector as in the second case. It is an eigenvector with eigenvalue -1, not 3.

[tex]\begin{bmatrix} 7 & 16 & 8 \\ -1 & 0 & -1 \\-2 & -10 & -3\end{bmatrix}\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix}23 \\ -1 \\ -12\end{bmatrix}[/tex]
so this is not an eigenvector.

[tex]\begin{bmatrix} 7 & 16 & 8 \\ -1 & 0 & -1 \\-2 & -10 & -3\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}= \begin{bmatrix}24 \\ -1 \\ -13\end{bmatrix}[/tex]
so this is not an eigenvector.
 
Last edited by a moderator:
  • #4
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I did the work again and found my mistake. The eigenvectors are

[0 -1 2], [2 -1 1], [-1 0 1]
 

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